Solution for Question 1
Method 1 : Trial and improvement method
a.) (by substituting the value of r )
If r= 1, A=
If r=2, A=
.…..
……
.…..
Surface area is minimum, when r = 5.4 , h = , and h/r =
b.) Alternative method , by substituting the value of h
Method 2 : Calculus ( Differentiation)
Method 3 : Graphical Method
By plotting the surface area A, against radius , r, for which A = ,
Construct table
r =1, 2, 3, ....8, 9 , A =
r =5.1, 5.2,...5.9, A =
r =
From the graph, the value of r =
h =
h/r =
Surface area is minimum, when r = 5.4 , h = , h/r =
To verify the answer, the following method can be used:
Construct table between r and A, r = 5 , 5.05, 5.1, ………, 5.65. Identified the minimum value of A. The table will shows that the minimum surface area occurs when the value of r falls between 5.4 and 5.5. Therefore the minimum surface area lies between 553.5881 and 553.7027.
To find the more precise value of r and A, the values of r from 5.41 to 5.54 are used.
From the table, the least value of surface area is 553.5811 and it occurs when r = 5.42
To find the more precise value of r and A, the values of r from 5.411 to 5.424 are used.
From the table, the least value fro surface area is 553.581 and it occurs when r = 5.419
Solution for Question 2
Method 1 : Calculus ( differentiation )
Method 2 : Draw graph
By plotting the surface area A, against radius r , for which A=
It can also be verify by the same method on question 1
Solution for Question 3
Consider the following shapes
i.) Equilateral Triangle
Method 1 : trigonometry and calculus (differentiation)
Method 2 : graphical method ( area versus radius)
Method 3 : Construct table ( verify more precise value)
ii.) Regular hexagon
iii.)
iv.)
Additional Task
o Consider a regular polygon with n sides
o Use calculus (differentiation)
o Generate table using excel.
Conclusion
…………………
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